Momentum and Newton's laws of motion

Calculate the momentum of:
a. A lab trolley of mass 1.0 kg moving at 20 cm.s^-1
b. A car of mass 650 kg moving at 24 m.s^-1
c. The Earth, mass 6.0 × 10²⁴ kg, moving at 29.8 km.s^-1 in its orbit around the Sun.

When calculating momentum, we must convert the unit of velocity into m.s^-1 and the unit of mass into kg

a. `20 " "cm.s^-1= 20xx0.01=0.2 " "m.s^-1`

`"momentum"=0.2" "m.s^-1xx "1 kg"=0.2" "kg.m.s^-1`

b. 

`"momentum"=24" "m.s^-1xx "650 kg"=15600" "kg.m.s^-1`

c. `29.8km.s^-1= 29.8xx1000=29800 m.s^-1`

`"momentum"=29800" "m.s^-1xx (6xx10^24 " "kg)=1.8xx10^29" "kg.m.s^-1`

A car of mass 750 kg accelerates from 10 m.s^-1 to 25 m.s^-1 in a time of 22.5 s.
a. Calculate the change in momentum of the car.
b. Use your answer to part a. to calculate the force causing the car to accelerate.

a. `"change in momentum" ="750 kg"xx(25" "m.s^-1-10" "m.s^-1)=11250" "kg.m.s^-1`

b. `"force"="change in momentum"/"time taken"=(11250" "kg.m.s^-1)/"22.5 s"="500 N"`

Use the idea of inertia to explain why some large cars have power-assisted brakes.

The greater the mass of the car, the greater the force needed to slow it down with a given deceleration. For large cars, it is less demanding on the driver if the engine supplies some of the force needed to brake the car.

A car crashes head-on into a brick wall. Use the idea of inertia to explain why the driver is more likely to come out through the windscreen if he or she is not wearing a seat belt.

Due to inertia, the driver continues to move forward, although the car stops. A seat belt provides the force needed to overcome this inertia.

A car starts to move along a straight, level road. For the first 10 s, the driver maintains a constant acceleration of 1.5 m.s^-2. The mass of the car is 1100 kg.
a. Calculate the driving force provided by the wheels, when:

i. The force opposing motion is negligible
ii. The total force opposing the motion of the car is 600 N.

b. Calculate the distance travelled by the car in the first 10 s.

a.

i. `"resultant force" =mxxa="1100 kg" xx1.5" " m.s^-2="1650 N"`

In this case, the driving force​ is equal to the resultant force since there is no opposing force.

Hence, `"driving force" " = " "1650 N"`

ii.

The total force opposing the motion is 600 N, so the driving force will be the net force plus the opposing force.

`"driving force"="resultant force"+"opposing force" = "1650 N" + "600 N"="2250 N"`

b. 

`s=ut+1/2at^2=0+1/2xx1.5" "m.s^2xx(10" "s)^2="75 m"`

A submarine of total mass 3200 kg is at rest underwater.

The total mass of the submarine is suddenly decreased by 200 kg by pumping water out of the submarine horizontally in a negligible time. The upthrust acting on the submarine is unchanged. The change in the total weight of the submarine causes it to accelerate vertically upwards. What is the initial upwards acceleration of the submarine?

Initial weight of the submarine:

`W_1="3200 kg" xx9.81" "m.s^-2="31392 N"`

Final weight of the submarine (after pumping out water):

`W_2="3000 kg" xx9.81" "m.s^-2="29430 N"`

The net force acting on the submarine is the difference between the upthrust and the final weight

`F_("net")=F_(upthrust)-W_2`

When the submarine's mass is reduced by pumping water out (removing 200 kg of mass), the weight of the submarine decreases. However, the upthrust remains the same because the submarine is still submerged in the water and the buoyant force depends on the volume of the submarine submerged.

`F_("net")=W_1-W_2="1952 N"`

`"accerleration"=F_("net")/"mass"="1952 N"/"3000 kg"=0.654" "m.s^-2`

This question is about Newton’s third law of motion. 

a. Two bar magnets are placed close to one another with their north poles facing each other.
i. State whether the magnets attract or repel each other.
ii. What does Newton’s third law tell you about the force each magnet exerts on the other?
b. If you stand on the floor, two forces act on you: your weight and the upward contact force of the floor.
i. Explain why these two forces are not an ‘equal and opposite pair’ in the sense of Newton’s third law.
ii. For each of the two forces, state the force that is equal and opposite to it, as described by Newton’s third law. Remember that ‘weight’ is the Earth’s gravitational pull on an object.

a.

i. Repel each other

ii.  The forces are equal in magnitude and opposite in direction. They act on different objects (the two magnets)

b. 

i.  The two forces act on the same object, not on different objects.

ii.  The ‘pair’ to weight is the gravitational pull of the person on the Earth. The ‘pair’ to the contact force of the floor on the person is the contact force the person exerts on the floor.

A parachutist has a mass of 95 kg. She is acted on by an upward force of 1200 N caused by her parachute. (acceleration due to gravity g = 9.81 m.s^-2) 

a. Calculate the parachutist’s weight.
b. Calculate the resultant force acting on her and give its direction.
c. Calculate her acceleration and give its direction

a. `"weight"=mxxg="95 kg"xx9.81" "m.s^-2="932 N"`

b. The upward force of 1200 N due to the parachute

The downward force of 932 N due to the weight of the parachutist

Since the upward force is greater than the downward force, the resultant force will be upward

`F_("net")="1200 N"-"932 N"="268 N"`

c. `"acceleration"=(F_("net"))/"mass"="268 N"/"95 kg"=2.82" "m.s^-2`

The weights and masses of four different objects on the surfaces of four different planets are shown. Which planet has the lowest value of acceleration of free fall at its surface?

Answer: D

Planet A:

`"acceleration of free fall"=(40xx10^-3 " N ")/(6xx10^-3 " kg ")=6.67 " " m.s^-2`

Planet B:

`"acceleration of free fall"=(3 " N ")/(500xx10^-3 " kg ")=6 " " m.s^-2`

Planet C:

`"acceleration of free fall"=(10 " N ")/(1 " kg ")=10 " " m.s^-2`

Planet D:

`"acceleration of free fall"=(2.6xx10^3 " N ")/(750 " kg ")=3.47 " " m.s^-2`

An insect of mass 4.5 mg, flying with a speed of 0.12 m.s^-1, encounters a spider’s web, which brings it to rest in 2.0 ms. Calculate the average force exerted by the insect on the web.

The change in momentum of the insect:

`"change in momentum"=4.5xx10^-6" kg " xx(0" " m.s^-1 - 0.12" "m.s^-1)= -5.4xx10^-7 " " kg.m.s^-1`

The average force exerted by the insect on the web:

`"average force"="change in momentum"/"time taken"=(-5.4xx10^-7 " " kg.m.s^-1)/"2 s"=-2.7xx10^-7 " " N`

The force exerted by the insect on the web is equal in magnitude but opposite in direction to the force exerted by the web on the insect. Therefore, the average force exerted by the insect on the web is `2.7xx10^-7 " " N`